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36. (ITA 1992) Num triângulo \( A B C \), retângulo em \( \hat{A} \), temos \( \hat{B}=60^{\circ} \). As bissetrizes destes ângulos se encontram num ponto \( D \). Se o segmento de reta \( B D \) mede 1 cm , então a hipotenusa mede a) \( \frac{1+\sqrt{3}}{2} \mathrm{~cm} \). b) \( 1+\sqrt{3} \mathrm{~cm} \). c) \( 2+\sqrt{3} \mathrm{~cm} \). d) \( 1+2 \sqrt{2} \mathrm{~cm} \). e) nenhuma das alternativas.

Ask by Mccoy Sullivan. in Brazil
Mar 15,2025

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A hipotenusa mede \(1 + \sqrt{3}\) cm.

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Para resolver esse problema, primeiro precisamos entender as relações de ângulos e os lados do triângulo. No triângulo \(ABC\), que é retângulo em \(A\), temos \( \hat{B} = 60^\circ \) e \( \hat{C} = 30^\circ \). A bissetriz do ângulo \(B\) (60º) divide-o em dois ângulos de 30º e se encontra com a bissetriz do ângulo \(C\) (30º) em \(D\). Sabemos que em um triângulo retângulo, a relação entre as medidas dos lados pode ser expressa pela função tangente. Nesse caso, temos: \[ \tan(30^\circ) = \frac{\text{cateto oposto}}{\text{cateto adjacente}} = \frac{BD}{AD} \Rightarrow \frac{1}{\sqrt{3}} = \frac{1}{AD}. \] Assim, temos \(AD = \sqrt{3}\). Usando o teorema de Pitágoras para calcular a hipotenusa \(AC\): \[ AC = \sqrt{AB^2 + BC^2} = \sqrt{(1 + \sqrt{3})^2 + (1)^2} = 2 + \sqrt{3}. \] Assim, a hipotenusa mede \(2 + \sqrt{3}\) cm, que corresponde à alternativa c) \(2 + \sqrt{3} \mathrm{~cm}\).

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